Suppose T is a bounded operator on a Hilbert space. Show that if λ is in the residual spectrum of T, then $\bar{λ}$ is in the point spectrum of the adjoint.

Here is what I think needs to be done. We know that $\langle Tu,v\rangle = \langle u,T^*v\rangle = \overline{\langle T^*v,u\rangle}$. Does that help with connecting $\lambda$ with $\overline{\lambda}$?

  • $\begingroup$ < and > mean "less than" and "greater than". The correct symbols you mean are \langle and \rangle. $\endgroup$ Mar 17 '15 at 5:03
  • $\begingroup$ Okay, I will edit the question. $\endgroup$ Mar 17 '15 at 5:25
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    $\begingroup$ Maybe can help the fact $\overline{R(T-\lambda I)}=N(T^*-\overline{\lambda}I)^{\bot}$. Since $\lambda$ is in the residual spectrum of $T$, then $\overline{R(T-\lambda I)}\neq H$. And then exist an element $x\neq 0$ in $N(T^*-\overline{\lambda} I)$. $\endgroup$
    – MrSelberg
    Mar 17 '15 at 6:12
  • $\begingroup$ hm where does that fact come from? and what is R&N? $\endgroup$ Mar 17 '15 at 6:40
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    $\begingroup$ As Hee Kwon Lee pointed, $R$ is the range of the operator, $N$ is the Kernel. For a proof of this fact, see Functional Analysis, Rudin, Theorem 4.12 (page 94). $\endgroup$
    – MrSelberg
    Mar 17 '15 at 7:02

I will explain Selberg's solution : Since $R:=\overline{R(T-\lambda I)}$, i.e., range of $T-\lambda I$, is a closed subspace and $H\neq R$ (since $\lambda$ is a residual spectrum), then there exists $$ x_0\neq 0 \in V,\ H=R\oplus V $$

So $$ 0= \langle (T-\lambda I)x,x_0\rangle=\langle x, (T^\ast -\overline{\lambda } I ) x_0 \rangle $$

Since $x$ is arbitrary, $(T^\ast -\overline{\lambda } I ) x_0 =0$.

  • $\begingroup$ I see. Can someone explain why we use the kernel? Also, can someone suggest a good resource for learning about spectrum? $\endgroup$ Mar 17 '15 at 7:05
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    $\begingroup$ @BobbyJones We want to see the kernel because we want to prove something about it: we want to prove that the kernel is not trivial. Hence I thought in link the range of $T-\lambda I$ (the information we know) and the Kernel of $T^*-\bar{\lambda}I$. (bad english :( ) $\endgroup$
    – MrSelberg
    Mar 17 '15 at 7:09

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